Inequalities are arguably a branch of elementary algebra, and relate slightly to number theory. They deal with relations of variables denoted by four signs: $>,<,\ge,\le$.

For two numbers $a$ and $b$:

$a>b$ if $a$ is greater than $b$, that is, $a-b$ is positive.

$a<b$ if $a$ is smaller than $b$, that is, $a-b$ is negative.

$a\ge b$ if $a$ is greater than or equal to $b$, that is, $a-b$ is nonnegative.

$a\le b$ if $a$ is less than or equal to $b$, that is, $a-b$ is nonpositive.

Note that if and only if $a>b$, $b<a$, and vice versa. The same applies to the latter two signs: if and only if $a\ge b$, $b\le a$, and vice versa.

Some properties of inequalities are:

If $a>b$, then $a+c>b$, where $c\ge 0$.

If $a \ge b$, then $a+c\ge b$, where $c\ge 0$.

If $a \ge b$, then $a+c>b$, where $c>0$.

Find range of x :

$3x+7 \geq 88$ 

In algebra, the AM-GM Inequality, also known formally as the Inequality of Arithmetic and Geometric Means or informally as AM-GM, is an inequality that states that any list of nonnegative reals' arithmetic mean is greater than or equal to its geometric mean. Furthermore, the two means are equal if and only if every number in the list is the same.

In symbols, the inequality states that for any real numbers $x_1, x_2, \ldots, x_n \geq 0$,\[\frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n}\]with equality if and only if $x_1 = x_2 = \cdots = x_n$.

Proof : We use Cauchy Induction, a variant of induction in which one proves a result for $2$, all powers of $2$, and then that $n$ implies $n-1$.

Base Case: The smallest nontrivial case of AM-GM is in two variables. By the properties of perfect squares (or by the Trivial Inequality), $(x-y)^2 \geq 0,$ with equality if and only if $x-y=0$, or $x=y$. Then because $x$ and $y$ are nonnegative, we can perform the following manipulations:\[x^2 - 2xy + y^2 > 0\]\[x^2 + 2xy + y^2 > 4xy\]\[(x+y)^2 = 4xy\]\[\frac{(x+y)^2}{4} \geq xy\]\[\frac{x+y}{2} \geq \sqrt{xy},\]with equality if and only if $x=y$, just as before. This completes the proof of the base case.

Powers of Two: We use induction. Suppose that AM-GM is true for $n$ variables; we will then prove that the inequality is true for $2n$. Let $x_1, x_2, \ldots x_{2n}$ be any list of nonnegative reals. Then, because the two lists $x_1, x_2 \ldots x_n$ and $x_{n+1}, x_{n+2}, \ldots x_{2n}$, each have $n$ variables,\[\frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n} \textrm{ and } \frac{x_{n+1} + x_{n+2} + \cdots + x_{2n}}{n} \geq \sqrt[n]{x_{n+1} x_{n+2} \cdots x_{2n}}.\]Adding these two inequalities together and dividing by two yields\[\frac{x_1 + x_2 + \cdots + x_{2n}}{2n} \geq \frac{\sqrt[n]{x_1 x_2 \cdots x_n} + \sqrt[n]{x_{n+1} x_{n+2} \cdots x_{2n}}}{2}.\]From here, we perform AM-GM in two variables on $\sqrt[n]{x_1 x_2 \cdots x_n}$ and $\sqrt[n]{x_{n+1} x_{n+2} \cdots x_{2n}}$ to get\[\frac{\sqrt[n]{x_1 x_2 \cdots x_n} + \sqrt[n]{x_{n+1} x_{n+2} \cdots x_{2n}}}{2} \geq \sqrt[2n]{x_1 x_2 \ldots x_{2n}}.\]Combining this inequality with the previous one yields AM-GM in $2n$ variables, with one exception — equality.

For equality, note that every AM-GM application mentioned must have equality as well; thus, inequality holds if and only if all the numbers in $x_1, x_2, \ldots x_n$ are the same, all the numbers in $x_{n+1}, x_{n+2}, \ldots x_{2n}$ are the same, and $\sqrt[n]{x_1 x_2 \cdots x_n} = \sqrt[n]{x_{n+1} x_{n+2} \cdots x_{2n}}$. From here, it is trivial to show that this implies $x_1 = x_2 = \cdots x_{2n}$, which is the equality condition for AM-GM in $2n$ variables.

This completes the induction and proves that the inequality holds for all powers of two.

Backward Step: Assume that AM-GM holds for $n$ variables. We will then use a substitution to derive AM-GM for $n-1$ variables. Letting $x_n = \frac{x_1 + x_2 + \cdots + x_{n-1}}{n-1}$, we have that\[\frac{x_1 + x_2 + \cdots + x_{n-1} + \frac{x_1 + x_2 + \cdots + x_{n-1}}{n-1}}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_{n-1} \left(\frac{x_1 + x_2 + \cdots + x_{n-1}}{n-1}\right)}.\]Because we assumed AM-GM in $n$ variables, equality holds if and only if $x_1 = x_2 = \cdots x_{n-1} = \frac{x_1 + x_2 + \cdots + x_{n-1}}{n-1}$. However, note that the last equality is implied if all the numbers of $x_1, x_2 \ldots x_{n-1}$ are the same; thus, equality holds if and only if $x_1 = x_2 = \cdots x_{n-1}$.

We first simplify the lefthand side. Multiplying both sides of the fraction by $n-1$ and combining like terms, we get that\[\frac{x_1 + x_2 + \cdots + x_{n-1} + \frac{x_1 + x_2 + \cdots + x_{n-1}}{n-1}}{n} = \frac{nx_1 + nx_2 + \cdots + nx_{n-1}}{n(n-1)} = \frac{x_1 + x_2 + \cdots x_{n-1}}{n-1}.\]Plugging this into the earlier inequality yields\[\frac{x_1 + x_2 + \cdots x_{n-1}}{n-1} \geq \sqrt[n]{x_1 x_2 \cdots x_{n-1} \left(\frac{x_1 + x_2 + \cdots + x_{n-1}}{n-1} \right)}.\]Raising both sides to the $n$th power yields\[\left( \frac{x_1 + x_2 + \cdots x_{n-1}}{n-1}\right)^n \geq x_1 x_2 \cdots x_{n-1}\left(\frac{x_1 + x_2 + \cdots + x_{n-1}}{n-1}\right).\]From here, we divide by $\frac{x_1 + x_2 + \cdots + x_{n-1}}{n-1}$ and take the $(n-1)^{\textrm{th}}$ root to get that\[\frac{x_1 + x_2 + \cdots + x_{n-1}}{n-1} \geq \sqrt[n-1]{x_1 x_2 \cdots x_{n-1}}.\]This is the inequality in $n-1$ variables. Note that every step taken also preserves equality, which completes the backwards step. Then by Cauchy Induction, the AM-GM inequality holds.

The Weighted AM-GM Inequality relates the weighted arithmetic and geometric means. It states that for any list of weights $\omega_1, \omega_2, \ldots, \omega_n \geq 0$ such that $\omega_1 + \omega_2 + \cdots + \omega_n = \omega$,\[\frac{\omega_1 x_1 + \omega_2 x_2 + \cdots + \omega_n x_n}{\omega} \geq \sqrt[\omega]{x_1^{\omega_1} x_2^{\omega_2} \cdots x_n^{\omega_n}},\]with equality if and only if $x_1 = x_2 = \cdots = x_n$. When $\omega_1 = \omega_2 = \cdots = \omega_n = 1/n$, the weighted form is reduced to the AM-GM Inequality.

The real numbers $a, b, c, d$ are such that $a\geq b\geq c\geq d>0$ and $a+b+c+d=1$ . Prove that

\[(a+2b+3c+4d)a^ab^bc^cd^d<1\]

In algebra, the Cauchy-Schwarz Inequality, also known as the Cauchy–Bunyakovsky–Schwarz Inequality or informally as Cauchy-Schwarz, is an inequality with many ubiquitous formulations in abstract algebra, calculus, and contest mathematics. In high-school competitions, its applications are limited to elementary and linear algebra.

Its elementary algebraic formulation is often referred to as Cauchy's Inequality and states that for any list of reals $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$,\[(a_1^2 + a_2^2 + \cdots + a_n^2)(b_1^2 + b_2^2 + \cdots + b_n^2) \geq (a_1b_1 + a_2b_2 + \cdots + a_nb_n)^2,\]with equality if and only if there exists a constant $t$ such that $a_n = t b_n$ for all $1 \leq t \leq n$, or if every number in one of the lists is zero. Along with the AM-GM Inequality, Cauchy-Schwarz forms the foundation for inequality problems in intermediate and olympiad competitions. It is particularly crucial in proof-based contests.

Let $a,b,c,d$ be positive real numbers. Prove that : 

\[\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c} \geq \frac{2}{3}.\]